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Question

Show that the force on each plate of a parallel plate capacitor has a magnitude equal to (12) QE, where Q is the charge on the capacitor, and E is the magnitude of electric field between the plates. Explain the origin of the factor .

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Solution

Let F be the force applied to separate the plates of a parallel plate capacitor by a distance of x. Hence, work done by the force to do so =Fx

As a result, the potential energy of the capacitor increases by an amount given as uAx

Where,

u= Energy density

A= Area of each plate

d= Distance between the plates

V= Potential difference across the plates

The work done will be equal to the increase in the potential energy i.e.,

Fx=uAx

F=uA=(120E2)A

Electric intensity is given by,

E=Vd

F=120(Vd)EA=12(0AVd)E

However, capacitance, C=0Ad

F=12(CV)E

Charge on the capacitor is given by,

Q=CV

F=12QE

The physical origin of the factor, 12, in the force formula lies in the fact that just outside the conductor, field is E and inside it is zero. Hence, it is the average value, E2, of the field that contributes to the force.


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